Work and Kinetic Energy Chapter 6 Work and Kinetic Energy

Understand & calculate work done by a force Goals for Chapter 6 Understand & calculate work done by a force Understand meaning of kinetic energy Learn how work changes kinetic energy of a body & how to use this principle Relate work and kinetic energy when forces are not constant or body follows curved path To solve problems involving power

Introduction The simple methods we’ve learned using Newton’s laws are inadequate when the forces are not constant. In this chapter, the introduction of the new concepts of work, energy, and the conservation of energy will allow us to deal with such problems.

Work A force on a body does work if the body undergoes a displacement.

Work done by a constant force The work done by a constant force acting at an angle  to the displacement is W = Fs cos .

Work done by a constant force The work done by a constant force acting at an angle  to the displacement is W = Fs cos . Units of Work = Force x Distance = Newtons x meters - (Nm) = JOULE of energy!

Positive, negative, and zero work A force can do positive, negative, or zero work depending on the angle between the force and the displacement.

Positive, negative, and zero work A force can do positive, negative, or zero work depending on the angle between the force and the displacement.

Positive, negative, and zero work A force can do positive, negative, or zero work depending on the angle between the force and the displacement.

Positive, negative, and zero work A force can do positive, negative, or zero work depending on the angle between the force and the displacement. Work is done BY an external force, ON an object. Positive work done by an external force

Positive, negative, and zero work A force can do positive, negative, or zero work depending on the angle between the force and the displacement. Work is done BY an external force, ON an object. Positive work done by an external force (with no other forces acting in that direction of motion)

Positive, negative, and zero work A force can do positive, negative, or zero work depending on the angle between the force and the displacement. Work is done BY an external force, ON an object. Positive work done by an external force (with no other forces acting in that direction of motion) will speed up an object!

Positive, negative, and zero work A force can do positive, negative, or zero work depending on the angle between the force and the displacement. Work is done BY an external force, ON an object. Positive work done by an external force (with no other forces acting in that direction of motion) will speed up an object! Negative work done by an external force (wnofaitdom) will slow down an object!

Work done by several forces – Example 6.2 Tractor pulls wood 20 m over level ground; Weight = 14,700 N; Tractor exerts 5000 N force at 36.9 degrees above horizontal. 3500 N friction opposes!

Work done by several forces – Example 6.2 How much work is done BY the tractor ON the sled with wood? How much work is done BY gravity ON the sled? How much work is done BY friction ON the sled?

Work done by several forces – Example 6.2 Tractor pulls wood 20 m over level ground; w = 14,700 N; Tractor exerts 5000 N force at 36.9 degrees above horizontal. 3500 N friction opposes!

Work done by several forces – Example 6.2 Work done BY tractor ON sled: = +5000N cos(36.9) * 20m = 80 kJ Work done BY gravity ON sled? = 14,700 N sin(90) * 20m = 0 Work done BY friction ON sled? = 3500N cos(180) * 20m = - 70 kJ NET work: +10 kJ

Gain KE doing POSITIVE WORK Kinetic energy The kinetic energy of a particle is KE = 1/2 mv2. NET work on body changes its kinetic energy! (& speed!) Gain KE doing POSITIVE WORK

Lose KE doing NEGATIVE WORK Kinetic energy The kinetic energy of a particle is K = 1/2 mv2. Net work on body changes its speed & kinetic energy Lose KE doing NEGATIVE WORK

The work-energy theorem The work done by the net force on a particle equals the change in the particle’s kinetic energy. Mathematically, the work-energy theorem is expressed as Wtotal = KEfinal – KEinitial = KE

Work done by several forces – Example 6.3 Tractor pulls wood 20 m over level ground; w = 14,700 N; Tractor exerts 5000 N force at 36.9 degrees above horizontal. 3500 N friction opposes! Suppose sled moves at 2.0 m/s; what is speed after 20 m?

Using work and energy to calculate speed Net gain in Work: +10kJ = CHANGE in KE KE initial = ½ mvi2 = ½ 1500 kg * 2.0m/s 2 = 3000 J KE final = ½ mvf2 = 3000 J + Net work done = 13 kJ Vf = 4.2 m/s! 1500 kg Net Work done = +10kJ

Forces on a hammerhead – example 6.4 Hammerhead of pile driver drives a beam into the ground falling along rails which provide 60N of frictional force. 200 kg head of pile driver lifted 3 m and dropped onto beam which sinks 7.4 cm into ground

Forces on a hammerhead – example 6.4 Hammerhead of pile driver drives a beam into the ground falling along rails which provide 60N of frictional force. Find speed as it hits and average force on the beam.

Forces on a hammerhead – example 6.4 Hammerhead of pile driver drives a beam into the ground falling along rails which provide 60N of frictional force. While falling down along rails… What does + work? What does – work?

Forces on a hammerhead – example 6.4 Hammerhead of pile driver drives a beam into the ground falling along rails which provide 60N of frictional force. How fast is it going just before hitting the beam? DKE = Net Work!

Forces on a hammerhead – example 6.4 Hammerhead of pile driver drives a beam into the ground falling along rails which provide 60N of frictional force. How fast is it going just before hitting the beam? DKE = Net Work! + 1900 N * 3 m = 5700 J Vf = 7.55 m/s

Forces on a hammerhead – example 6.4 Hammerhead of pile driver drives a beam into the ground, following guide rails that produce 60 N of frictional force While hitting the beam What does + work? What does – work? What distance do these forces act?

Forces on a hammerhead – example 6.4 Hammerhead of pile driver drives a beam into the ground, following guide rails that produce 60 N of frictional force While hitting the beam Know KEi = 5700 J! Know KEf = 0! Know distance = 7.4 cm Get Net Force acting!

Comparing kinetic energies – example 6.5 Two iceboats have different masses, m & 2m. Wind exerts same force; both start from rest.

Comparing kinetic energies – example 6.5 Two iceboats have different masses, m & 2m. Wind exerts same force; both start from rest. Which boat wins? Which boat crosses with the most KE?

Kinetic energy does not depend on the direction of motion. © 2016 Pearson Education Inc.

Kinetic energy increases linearly with the mass of the object. © 2016 Pearson Education Inc.

Kinetic energy increases with the square of the speed of the object. © 2016 Pearson Education Inc.

The work-energy theorem The work-energy theorem: The work done by the net force on a particle equals the change in the particle’s kinetic energy. © 2016 Pearson Education Inc.

Work and energy with varying forces Many forces, such as force to stretch a spring, are not constant.

Work and energy with varying forces Many forces, such as force of a stretched spring, are not constant. Suppose a particle moves along the x-axis from x1 to x2 under a VARYING force

Work and energy with varying forces Approximate work by dividing total displacement into many small segments.

Work and energy with varying forces Calculate approximate work done by variable force over many segments. Do this for each segment & add results for all segments. © 2016 Pearson Education Inc.

Work and energy with varying forces—Figure 6.16 Work = ∫ F∙dx An infinite summation of tiny rectangles!

Work and energy with varying forces—Figure 6.16 Work = ∫ F∙dx An infinite summation of tiny rectangles! Height: F(x) Width: dx Area: F(x)dx Total area: ∫ F∙dx

Work and energy with varying forces The work done by the force in the total displacement from x1 to x2 is the integral of Fx from x1 to x2: On a graph of force as a function of position, the total work done by the force is represented by the area under the curve between the initial and final positions.

Work done by a constant force © 2016 Pearson Education Inc.

Stretching a spring The force required to stretch a spring a distance x is proportional to x: Fx = kx k is the force constant (or spring constant) Units of k = Newtons/meter Large k = TIGHT spring Small k = L O O S E spring

Stretching a spring The force required to stretch a spring a distance x is proportional to x: Fx = kx k is the force constant (or spring constant) of the spring. Area under graph represents work done on the spring to stretch it a distance X: W = ½ kX2

Work done on a spring scale – example 6.6 A woman of 600 N weight compresses spring 1.0 cm. What is k and total work done BY her ON the spring?

Work done on a spring scale – example 6.6 A woman of 600 N weight compresses spring 1.0 cm. What is k and total work done?

Work done on a spring scale – example 6.6 A woman of 600 N weight compresses spring 1.0 cm. What is k and total work done BY her ON the spring? K = Fs/x = -600N / -0.010 m = + 6.0 x 104 N/m W = ½ kxf2 – ½ kxi2 = ½ (6.0 x 104N/m) (-0.010 m)2 = 3.0 J

Motion with a varying force An air-track glider mass 0.1 kg is attached to a spring of force constant 20 N/m, Glider starts at rest with spring compressed, then released. At the point where the spring is no longer compressed, it is moving with some speed v1.

Motion with a varying force The force on the glider is varying from maximum + to 0 to maximum negative, when it stops momentarily. Moving at 1.50 m/s to right when spring is unstretched. Find maximum distance moved if no friction, and if friction was present with mk = 0.47

Motion with a varying force glider mass 0.1 kg spring of force constant 20 N/m Moving at 1.50 m/s to right when spring is unstretched. Final velocity once spring stops glider = 0 We know 3 things! F/m = a; vi; vf Why can’t we use F = ma ?

Motion with a varying force An air-track glider is attached to a spring, so the force on the glider is varying. In general, if the force varies, using ENERGY will be an easier method than using forces! We know: Initial KE = ½ mv2 Final KE = ½ mvf2 = 0 Get net work done! Get ½ kx2

Motion with a varying force An air-track glider is attached to a spring, so the force on the glider is varying.

Motion on a curved path—Example 6.8 A child on a swing moves along a smooth curved path at constant speed. Weight w, Chain length = R, max angle = q0 You push with force F that varies. What is work done by you? What is work done by gravity? What is work done by chain?

Motion on a curved path—Example 6.8 A child on a swing moves along a curved path. Weight w, Chain length = R, max angle = q0 You push with force F that varies.

Motion on a curved path—Example 6.8 W = ∫ F∙dl F∙dl = F cos q dl = ds (distance along the arc)

Work–energy theorem for motion along a curve A particle moves along a curved path from point P1 to P2, acted on by a force that varies in magnitude and direction. © 2016 Pearson Education Inc.

Work–energy theorem for motion along a curve A particle moves along a curved path from point P1 to P2, acted on by a force that varies in magnitude and direction. The work can be found using a line integral: © 2016 Pearson Education Inc.

Power is the rate at which work is done. Average power is: Instantaneous power is: The SI unit of power is the watt (1 W = 1 J/s), but another familiar unit is the horsepower (1 hp = 746 W).

Power: Lifting a box slowly © 2016 Pearson Education Inc.

Power: Lifting a box quickly © 2016 Pearson Education Inc.

kilowatt-hour (kwh) is ENERGY (power x time) Power is rate work is done. Average power is Pav = W/t Instantaneous power is P = dW/dt. SI unit of power is watt (1 W = 1 J/s) horsepower (1 hp = 746 W ~ ¾ of kilowatt) kilowatt-hour (kwh) is ENERGY (power x time)

In mechanics we can also express power in terms of force and velocity: Here is a one-horsepower (746-W) propulsion system.

A “power climb” A person runs up stairs.

Banked Curve Examples Ny mg N Nx center of curvature of the banked road x y fsx fsy fs

Banked Curve Examples Maximum Cornering Speed for Banked Curves Radius (m) 30 40 50 60 Coefficient of Static Friction 0.8 0.6 0.4 0.2 Angle (degrees) 20 10 5   max speed (m/s) 27 22 18 15 23 19 16 21 11 max speed (mph) 48 34 51 43 35 41 46 24